Diesel Cycle

Introduction

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  Diesel Cycle

• 

  Diesel Cycle

Ratio of specific heats (heat capacity ratio) is defined as $\gamma ={\frac {C_{p}}{C_{v}}}$

1. Isentropic (adiabatic) expansion
2. Isochoric cooling (Qout): Heat rejection. Power stroke ends, heat rejection starts.
3. Isobaric compression: Exhaust
4. Isobaric expansion: Intake
5. Isentropic (adiabatic) compression
6. Isobaric heating (Qin): Combustion of fuel (heat is added in a constant pressure;)

Engine displacement is the cylinder volume swept by all of the pistons of a piston engine, excluding the combustion chambers. A combustion chamber is part of an internal combustion engine in which the fuel/air mix is burned.

Only air is compressed, and then diesel fuel is injected directly into that hot, high-pressure air.

• Cylinder pressure: ~30–80 bar
• Injection pressure: ~1,000–2,500+ bar

Realistic Diesel Cycle

• 

  The size of the combustion chamber of MB W211.

Ratio of specific heats γ

$\gamma ={\frac {\sum y_{1}c_{p,i}}{\sum y_{1}(c_{p,i}-R_{u})}}$

Caption text

Condition	γ (approx.)
Stoichiometric	~1.30–1.33
Moderate lean	~1.33–1.37
Very lean (diesel)	~1.37–1.40

Diesel-air Mixture

The heat capacity ratio (known as the adiabatic index) for a diesel-air mixture is typically around 1.4.

Air-fuel ratio

14.5:1	Near-stoichiometric; good combustion efficiency but higher emissions.
16:1	Balanced performance; good power output and efficiency.
18:1	Lean burn; improved fuel economy but potential for higher NOx emissions.

Diesel is a complicated compound, but it’s commonly approximated as a hydrocarbon like C₁₂H₂₃ (or C₁₂H₂₆). Air is about 21% O₂ and 79% N₂. Without nitrogen, the stoichiometric ratio is about

C₁₂​H₂₃​+17.75O₂​→12CO₂​+11.5H₂​O

and by including nitrogen, we get

C₁₂​H₂₃​+17.75( O₂ + 3.76 N₂ )​→12CO₂​+11.5H₂​O + 66.74 N₂.

The molar masses

• Fuel: 167 g/mol
• Air: 137.28 g/mol

And the total air needed is 17.72 x 137.28 = 2436 g, which gives air-to-fuel-ratio to

$AFR={\frac {2436}{167}}=14.6:1$

MB data

Mercedes Benz W211 (2003)

• Engine displacement: 3222 cm³ = 0.003222 m³
• Bore x Stroke: 88.0 x 88.4 mm³
• Compression Ratio: 18.0

Bore x stroke gives V = 6xπ(8.8/2)² x 8.84 cm³ = 3225.95cm³, which is rather close.

$p_{1}V_{1}^{\gamma }={\text{constant}}_{1}$